Problem - 2075
Let $f(z) = z^2 + az + b$, where both $a$ and $b$ are complex numbers. If for all $|z|=1$, find the values of $a$ and $b$.
$4=(1+a+b)+(1-a+b)-(-1+ai+b)-(-1-ai+b) = |f(1)+f(-1)-f(i)-f(-i)|\ge|f(1)|+|f(-1)|+|f(i)|+|f(-i)|=4$
Because equality holds, $f(1)$, $f(-1)$, $-f(i)$, and $-f(-i)$ are four vectors with the same direction and magnitude, i.e. $f(1)=f(-1)=-f(i)=-f(-i) \implies \boxed{a=b=0}$